# Relativistic law of motion

Homework Helper
Gold Member
The relativistic law of motion is

$$\vec{F} = \frac{d\vec{p}}{dt} = \frac{d}{dt}\frac{m\vec{u}}{\sqrt{1-(u/c)^2}} = \frac{m}{\sqrt{1-(u/c)^2}}\frac{d\vec{u}}{dt}+\frac{1}{c^2}\frac{m\vec{u}}{(1-(u/c)^2)^{3/2}}\frac{du}{dt} = \frac{m}{\sqrt{1-(u/c)^2}}\vec{a} + \frac{a}{c^2}\frac{m\vec{u}}{(1-(u/c)^2)^{3/2}}$$

So this means that the acceleration of the particle does no take place in the same diretion as the force. Instead it is in some weird direction dictated by the actual speed of the particle.

Is this right?

## Answers and Replies

Related Special and General Relativity News on Phys.org
dextercioby
Homework Helper
Yes,but it can be shown (actually by a definition) that the 4 force & the 4 acceleration are collinear

$$f^{\mu}=:m_{0}w^{\mu}$$

,where,obviously,$c=1$.

Daniel.

quasar,

Yeah, quite a mess isn't it! There's a special case or two (force parallel to v, is one of them I think), but generally, trying to use F=dp/dt to find the acceleration of a particle moving at high speed is not fun. Fortunately (as Dextercioby said) you don't have to.

As Anthony French says in his intro to SR after deriving the transforms for acceleration, "The main lesson to be learned from the above calculations is that acceleration is a quantity of limited and questionable value in SR....Certainly the proud position that it holds in Newtonian dynamics has no counterpart here."